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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\cl {\bf Free Rotational Motion of Rigid Bodies }
\lf
 What is to observe in the 3D-XplorMath exhibit \lf
 \Hskip1.5cm {\it Solid Body (Euler's Polhode)} ?
\vskip2mm\noindent
A brick -- in the program of edge lengths $aa\ge bb \ge cc \ge 0$ -- is a good example
of a solid (also: rigid) body. The program illustrates the free rotational movement of
such a brick (i.e. gravity is ignored): \lf
Select {\it Solid Body (Euler's Polhode)}, stop the alternation between two pictures by a 
mouse click and select {\it Do Poinsot Construction From Polhode} at
the bottom of the Action Menu. 
The resulting animation shows a freely tumbling brick. 
By changing $aa, bb, $ or $cc$ one may watch other bricks tumbling.
\lf
There are three other input parameters, $dd, ee, ff$. These are initial conditions for the
the tumbling motion. If one chooses 
$(dd,ee,ff)\approx (1,0.1,0.1)$ or $(dd,ee,ff)\approx (0.1,0.1,1)$
then there is not much tumbling. These motions are almost rotations around the longest 
axis (aa) of the brick, respectively the shortest axis (cc) .  The fact that these rotation
axes stay close to their initial position is expressed by saying: 
{\it the rotations around the longest and the shortest axis are stable}. 
Now look again at the default initial conditions
$(dd,ee,ff)\approx (0.1,1,0.1)$. One observes that the momentary axis of rotation moves
almost to the direction opposite to the initial direction and then returns back. One says:
{\it the rotation around the middle axis of the brick is  unstable}. -- By putting a tape 
around a book and trying to throw it so that it rotates around one of the three axes
one can experimentally test these theoretical predictions.
\LF
The {\it explanation} of this behaviour has a mathematical part and a physical part. The
physical part is contained in the initial picture, the mathematical part is the connection
between the initial picture and the annimation. We explain the mathematical part in
\vskip2mm\noindent
\cl { Part I: From Angular Velocity to Rotational Motion}
\vskip1.5mm\noindent
It is available in the {\it Topics} part of the {\it Documentation}. This mathematical part
has no physical limitations, any of the space curves in the program can be used as
{\it angular velocity curve} and in the Action Menu one can select animations that 
show the resulting motions.
\lf
The physical part requires in addition to {\it angular velocity} the physical notions
{\it tensor of inertia} and {\it angular momentum}. These are explained below. 
What can one say before this theory about the initial picture of the program?
We see two space curves. The one on the sphere is the angular momentum as a
function of  time {\it in the coordinate system of the brick}. The other one is the
angular velocity curve (called {\it Polhode}). Both are intersections of quadratic
surfaces, represented by dots in the picture. The two curves are related by a fixed
linear map -- given by the tensor of inertia. To emphasize this linear map the quadratic
surfaces alternate between the domain and the range of this map. Finally, these two
curves together determine Euler's differential equation for either of them. For example
the derivative of the angular momentum curve is the cross product of the corresponding
position vectors of the angular momentum curve and the angular velocity curve, in
formulas: $\vec\ell'(t) = \vec\ell(t) \times \vec\omega(t)$. The Action Menu entry
{\it Show Rep\`ere Mobile and ODE} illustrates this connection.  The dotted curves
on the sphere are solutions for other initial conditions $dd,ee,ff$ with the same value
$dd^2+ee^2+ff^2$. The default morph varies $bb$ between $aa$ and $cc$, it 
illustrates how the {\it family} of polhodes depends on the shape of the brick.
\lf
And here is the theory:
\vskip2mm\noindent
\cl { Part II: Tensor of Inertia and Angular Momentum  }
\vskip2mm\noindent
The tensor of inertia is a map that transforms {\it angular velocity} into {\it angular
momentum}.\lf
Historical note: The word {\it tensor} is a generic word that describes objects from linear
algebra that can be given by components (indices!) with respect to a base. 
The {\it tensor of inertia} is a {\it linear map} from the 3-dim vector space of angular velocities
to the 3-dim vector space of angular momenta. What we need below is that for each solid
body there exists an orthonormal frame $\{\vec e_x(t),\ \vec e_y(t),\ \vec e_z(t)\}$ in the rest
space of the body (i.e. moving with the body) so that the tensor of inertia $\Theta$ is a
diagonal map:
$$\eqalign{
&angular\ momentum = \Theta(\vec\omega(t)) =    \cr
&\omega_x(t)\cdot\Theta_x\vec e_x(t) + \omega_y(t)\cdot\Theta_y\vec e_y(t) 
+ \omega_z(t)\cdot\Theta_z\vec e_z(t).
}$$
$\Theta_x, \Theta_y, \Theta_z$ are called  {\it principal moments of inertia}.
\LF
We now explain the tensor of inertia in some more detail. The result of the explanation
will be the above formula for the angular momentum. We view a solid body as a collection
of points of {\bf mass} $m_i$ and position vector $\vec x_i(t)$; the pairwise distances
between these points are constant. The origin is the center of mass of these points, i.e.
$\sum_i m_i \vec x_i(t) =\vec 0$. For each mass point we have the following definitions,
the corresponding notions for the solid body are obtained by summation: \lf
linear momentum: $\vec p_i(t) := m_i\vec x_i\,'(t)$  \lf
angular momentum with respect to the origin: 
\lf \Hskip2.5cm$\vec\ell_i(t) := \vec x_i(t) \times \vec p_i(t) $ \lf
kinetic energy: $E_i(t) := {1\over 2}m_i\langle \vec x_i\,'(t) , \vec x_i\,'(t) \rangle$. \lf
The body is rigid, i.e.  the distances between the points are constant, therefore there is
an angular velocity function $\vec\omega(t)$ that relates the positions and velocities: \lf
rotational motion: $\vec x_i\,'(t) = \vec\omega(t)\times \vec x_i(t)$.  \lf
angular momentum: $\vec\ell_i(t) = \vec x_i(t) \times (\vec\omega(t)\times \vec x_i(t))$.  \lf
\phantom{angular momentum: $\vec\ell_i(t)$ }$ =: \Theta_i(\vec\omega(t))$.  \lf
This tensor of inertia is most easily understood if we use the relation between cross-product
and matrix-product and insert it into the above definitions. We obtain the expressions for
angular momentum and kinetic energy in terms of the tensor of inertia and the angular
velocity as follows:
\goodbreak
$$\eqalign{
 \vec\omega\times \vec x &= \left( \matrix{0 &-\omega_z &\omega_y \cr
                                                 \omega_z &0 &-\omega_x         \cr
                                                 -\omega_y &\omega_x &0
 } \right)\cdot  \left( \matrix{x \cr  y \cr z}\right)  \cr
 &=   \left( \matrix{0 &z &-y \cr
                              -z &0 &x         \cr
                               y &-x &0
 } \right)  \cdot \left( \matrix{\omega_x \cr  \omega_y \cr \omega_z}\right) 
}$$
We obtain
$$\eqalign{
&\vec\ell_i(t) =  \cr
&m_i\left( \matrix{0 &z_i &-y_i \cr
                         -z_i &0 &x_i         \cr
                           y_i &-x_i &0
 } \right)   \left( \matrix{0 &-z_i &y_i \cr
                                                z_i &0 &-x_i         \cr
                                                -y_i &x_i &0
 } \right)   \left( \matrix{\omega_x \cr  \omega_y \cr \omega_z}\right)    \cr
 &= m_i\left( \matrix{y_i ^2+ z_i^2&-x_iy_i &-x_iz_i \cr
                                               -x_iy_i  &y_i ^2+ z_i^2 &-y_iz_i          \cr
                                                -x_iz_i &-y_iz_i  &x_i ^2+ y_i^2
 } \right)   \left( \matrix{\omega_x \cr  \omega_y \cr \omega_z}\right) \cr
&= \Theta_i(\vec\omega)\ (\hbox{Note the symmetry of the matrix of } \Theta_i).
\cr
&E_i(t) = {1\over2}\langle  \Theta_i(\vec\omega),\  \vec\omega \rangle.
}$$
The symmetry of $\Theta:=\sum_i\Theta_i$ implies that we have an orthonormal
eigen basis for $\Theta$. The corresponding eigen values are the principal moments
of inertia, $\Theta_x, \Theta_y, \Theta_z$.
\goodbreak\noindent
Finally, we will derive Euler's equations, a first order ODE for $\vec\omega(t)$. Together
with part I this determines the motion of a solid body that rotates without exterior forces.
We will always take the eigen basis of $\Theta$ as the moving frame of part I.
\LF
Newton's laws imply that the total angular momentum is constant in situations that are
more general than the force free rotation of a solid body. We omit this general theory
and show only that the conservation of angular momentum is equivalent to Euler's
equations.
$$
\vec\ell(t) := \sum_i\vec\ell_i(t) = \Theta(\vec\omega(t)) = 
\sum_{\xi\in\{x,y,z\}}\omega_\xi(t)\Theta_\xi\vec e_\xi(t)
$$
implies
$$\eqalign{
&{d\over dt}\vec\ell(t) = \cr 
&\sum_{\xi\in\{x,y,z\}}\omega_\xi(t)\,'\Theta_\xi\vec e_\xi(t) +
            \sum_{\xi\in\{x,y,z\}}\omega_\xi(t)\Theta_\xi\vec e_\xi\,'(t).            
}$$
Insert $\vec e_\xi\,'(t) = \vec\omega(t)\times\vec e_\xi(t)$ to get
$$
\sum_{\xi\in\{x,y,z\}}\omega_\xi(t)\Theta_\xi\vec e_\xi\,'(t)=
\vec\omega(t)\times\vec\ell(t),
$$
next compute the cross product in the base given by the moving frame:
$$
\vec\omega(t)\times\vec\ell(t)= \sum_{\xi\in\{x,y,z\}}\left(
               \left(\matrix{\omega_x \cr  \omega_y \cr  \omega_z}\right)\times
               \left(\matrix{\ell_x \cr  \ell_y \cr  \ell_z}\right)\right)_\xi\cdot\vec e_\xi(t),
$$
finally compare coefficients to get Euler's equations:

\vskip1mm\noindent
{  \hbox{
       \vrule  \vbox{\hrule \hrule  
              \vbox{   \vskip2mm\noindent 
$$\eqalign{
   &\left(\matrix{\ell_x \cr  \ell_y \cr  \ell_z}\right)' = 
   \left(\matrix{\Theta_x\omega_x \cr \Theta_y \omega_y \cr  \Theta_z\omega_z}\right)'=
  - \left(\matrix{\omega_x \cr  \omega_y \cr  \omega_z}\right)\times
               \left(\matrix{\ell_x \cr  \ell_y \cr  \ell_z}\right), \ \ \ 
   \cr
   &\hbox{where the physics is contained in the relation}\cr
   &\hbox{between } \omega \hbox{ and } \ell:
   \cr
   &\ell_x = \Theta_x\omega_x,\  \ell_y = \Theta_y\omega_y,\ 
    \ell_z = \Theta_z\omega_z.
}$$}
                            \hrule \hrule} 
       \hskip-1.1mm  \vrule    }}
       
\noindent
Considered as differential equation for the $\omega$-compo\-nents these are Euler's
equations. This ODE-system implies immediately that the two quadratic functions
$$\eqalign{
  &|\vec{\ell}\,|^2 = \ell_x^2+\ell_y^2+\ell_z^2 =
  \Theta_x^2\omega_x^2+\Theta_y^2\omega_y^2+\Theta_z^2\omega_z^2 
  \hskip4mm\hbox{ and}  \cr
  &2E = \ell_x\omega_x+\ell_y\omega_y+\ell_z\omega_z =
   \Theta_x\omega_x^2+\Theta_y\omega_y^2+\Theta_z\omega_z^2
}$$
are constant along solution curves. The solutions are therefore intersections of
two ellipsoids. If one considers the ODE-system as differential equations for the
$\ell$-components then one of the ellipsoids is a sphere, the solutions
$(\ell_x(t), \ell_y(t), \ell_z(t))$ are spherical curves. The choice of the  $\ell$-components
as the functions to be determined therefore simplifies the visualization and also
leads to a slightly simpler ODE-system, since the tensor of inertia enters only on the
right side, linearly,  into the equations.




\bye
 